- #26

- 332

- 191

We have,

$$

I=\int_0^{\infty}\int_0^{\infty}e^{-(x+y+\frac{\lambda^3}{xy})}x^{-\frac{2}{3}}y^{-\frac{1}{3}}dxdy

$$

Make the following expansion under the integral sign,

$$

e^{-\frac{\lambda^3}{xy}}=\sum_{n=0}^{\infty}(\frac{\lambda^3}{xy})^n\frac{(-1)^n}{n!}

$$

to get

$$

I=\int_0^{\infty}e^{-y}y^{-\frac{1}{3}}\int_0^{\infty}e^{-x}x^{-\frac{2}{3}}(\sum_{n=0}^{\infty}(\frac{\lambda^3}{xy})^n\frac{(-1)^n}{n!})dxdy

$$

$$

=\int_0^{\infty}e^{-y}y^{-\frac{1}{3}}\sum_{n=0}^{\infty}(\int_0^{\infty}(-1)^n(\frac{\lambda^3}{y})^n\frac{x^{-n-\frac{2}{3}}}{n!}e^{-x}dx)dy\\

$$

Each x integral term in the expansion results in a Gamma function and taking the sum out side the remaining integral we get,

$$

I=\sum_{n=0}^{\infty}(-1)^n\lambda^{3n}\frac{\Gamma (\frac{1}{3}-n)}{n!}\int_0^{\infty}e^{-y}y^{-n-\frac{1}{3}}dy

$$

Again each term y integral term in the expansion results in a Gamma function.

$$

I=\sum_{n=0}^{\infty}(-1)^n\lambda^{3n}\Gamma (\frac{2}{3}-n)\frac{\Gamma (\frac{1}{3}-n)}{n!}

$$

By iterating each term in the sum using the identity,

$$

\Gamma (n-1)=\frac{\Gamma (n)}{n-1}

$$

we find,

$$

I=1+\sum_{n=1}^{\infty}(-1)^n\frac{\lambda^{3n}3^{2n}\Gamma (\frac{1}{3}) \Gamma (\frac{2}{3}) }{\Pi_{k=1}^n (3k-1) \Pi_{k=1}^n (3k-2)n!}

$$

Let ##x=3\lambda##. On expanding each term in the sum we find,

$$

I=\Gamma (\frac{1}{3})\Gamma (\frac{2}{3})\sum_{n=0}^{\infty}(-1)^n\frac{x^{3n}}{(3n)!}

$$

Let us seek a solution for ##y(x)=\sum_{n=0}^{\infty}(-1)^n\frac{x^{3n}}{(3n)!}## in the form of a second order differential equation. On expanding each term we have,

$$

y(x)= 1-\frac{x^3}{3!}+\frac{x^6}{6!}-\frac{x^9}{9!}+...

$$

$$

y'(x)=-\frac{x^2}{2!}+\frac{x^5}{5!}-\frac{x^8}{8!}+...

$$

$$

y''(x)=-x+\frac{x^4}{4!}-\frac{x^7}{7!}+\frac{x^{10}}{10!}-\frac{x^{13}}{13!}+...

$$

We observe,

$$

y''-y'+y=e^{-x}

$$

The characteristic polynomial for the homogeneous solution is,

$$

\lambda^2-\lambda + 1=0\\

$$

and thus,

$$

\lambda=\frac{1}{2}\pm \frac{i\sqrt{3}}{2}\\

$$

with the homogeneous solution,

$$

y_h(x)=e^{\frac{x}{2}}[c_1\cos(\frac{\sqrt{3}x}{2})+c_2\sin(\frac{\sqrt{3}x}{2})]

$$

The particular solution is of the form,

$$

y_p=Ce^{-x}

$$

Plugging this into the differential equation we find ##C=\frac{1}{3}##. The solution is the sum of the homogeneous part and the particular part,

$$

y(x)=y_h(x)+y_p(x)

$$

Applying the boundary conditions ##y(0)=1## and ##y'(0)=0## we find ##c_1=\frac{2}{3}## and ##c_2=0##. Thus

$$

y(x)=\frac{2}{3}e^{\frac{x}{2}}\cos(\frac{\sqrt{3}x}{2})+ \frac{1}{3}e^{-x}

$$

and we finally have,

$$

\int_0^{\infty}\int_0^{\infty}e^{-(x+y+\frac{\lambda^3}{xy})}x^{-\frac{2}{3}}y^{-\frac{1}{3}}dxdy=\Gamma (\frac{1}{3})\Gamma (\frac{2}{3})[\frac{2}{3}e^{\frac{(3\lambda)}{2}}\cos(\frac{3\sqrt{3}\lambda}{2})+ \frac{1}{3}e^{-3\lambda}]

$$